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  • Oracle查询易忘点

    1.查询字符‘截取’数据substr

    SELECT "age" SAGE, substr("name", 2, 2)  SNAME FROM Student//从字符下标为2开始截取,截取2个(首位下标为1,不是0)

    2.查询拼接无关联字段

     1 select sum(bdnrjk) as "bdnrjk",sum(bdnrck) as "bdnrck",sum(bddljk) as "bddljk",sum(bddlck) as "bddlck",sum(wdnrjk)as "wdnrjk",sum(wdnrck) as "wdnrck",sum(wddljk)as "wddljk",sum(wddlck) as "wddlck" from ( 
     2     select count(*) as bdnrjk, 0 as bdnrck , 0 as bddljk , 0 as bddlck , 0 as wdnrjk , 0 as wdnrck , 0 as wddljk , 0 as wddlck 
     3         from   TESTTABLE where 1=1 UNION all 
     4     select 0 as bdnrjk, count(*) as bdnrck , 0 as bddljk , 0 as bddlck , 0 as wdnrjk , 0 as wdnrck , 0 as wddljk , 0 as wddlck 
     5         from   TESTTABLE where 1=1 UNION all 
     6     select 0 as bdnrjk , 0 as bdnrck , count(*) as bddljk , 0 as bddlck , 0 as wdnrjk , 0 as wdnrck , 0 as wddljk , 0 as wddlck 
     7         from   TESTTABLE where 1=1 UNION all 
     8     select 0 as bdnrjk , 0 as bdnrck , 0 as bddljk , count(*) as bddlck , 0 as wdnrjk , 0 as wdnrck , 0 as wddljk , 0 as wddlck 
     9         from   TESTTABLE where 1=1 UNION all 
    10     select 0 as bdnrjk, 0 as bdnrck , 0 as bddljk , 0 as bddlck , count(*) as wdnrjk , 0 as wdnrck , 0 as wddljk , 0 as wddlck 
    11         from   TESTTABLE where 1=1 UNION all 
    12     select 0 as bdnrjk , 0 as bdnrck , 0 as bddljk , 0 as bddlck , 0 as wdnrjk , count(*) as wdnrck , 0 as wddljk , 0 as wddlck 
    13         from   TESTTABLE where 1=1 UNION all 
    14     select 0 as bdnrjk , 0 as bdnrck , 0 as bddljk , 0 as bddlck , 0 as wdnrjk , 0 as wdnrck , count(*) as wddljk , 0 as wddlck 
    15         from   TESTTABLE where 1=1 UNION all 
    16     select 0 as bdnrjk , 0 as bdnrck , 0 as bddljk , 0 as bddlck , 0 as wdnrjk , 0 as wdnrck , 0 as wddljk , count(*) as wddlck 
    17         from   TESTTABLE where 1=1
    18  ) 

    3.REGEXP_LIKE (字段,'值|值|值|值')

    select * from Student where REGEXP_LIKE (class,'1班|2班|3班|4班')//查询1、234、班的学生(即:class字段值为1班或2班或3班或4班的student)

     4.nvl(字段,默认值)

    select name,nvl(age,99) as AGE from Teacher//查询老师的名字和年龄,如果表中年龄字段为空则查出的数据会赋予默认值'99'

     5.CASE WHEN 条件 THEN 结果 ... ELSE 结果   END

     1 -- 2 select name
     3 CASE 
     4     when age=100 then '老师'
     5     when name='张三' then '三儿'
     6     when name='李四' then '四儿'
     7     else name
     8 end
     9 from Student
    10 --11 DELETE Student where 
    12 name=(case 
    13     when age=100 then '老师'
    14     when name='张三' then '三儿'
    15     when name='李四' then '四儿'
    16     else name
    17 end)
    18 
    19 --20 update Student set 
    21 name=(case 
    22     when age=100 then '老师'
    23     when name='张三' then '三儿'
    24     when name='李四' then '四儿'
    25     else name
    26 end)
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  • 原文地址:https://www.cnblogs.com/janesyf/p/7942350.html
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